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Status: Verified (1)

RFC 6234, "US Secure Hash Algorithms (SHA and SHA-based HMAC and HKDF)", May 2011

Source of RFC: IETF - NON WORKING GROUP
Area Assignment: sec

Errata ID: 5179
Status: Verified
Type: Editorial
Publication Format(s) : TEXT
Reported By: Russ Housley
Date Reported: 2017-11-06
Verifier Name: Roman Danyliw
Date Verified: 2022-01-19

Section 8.5 says: 

 *    This file will exercise the HKDF code performing
 *      the seven tests documented in RFC 4869.

It should say:

 *    This file will exercise the HKDF code performing
 *      the seven tests documented in RFC 5869.

Notes:

Typo. Provide the correct reference in the code comment.

Status: Reported (2)

RFC 6234, "US Secure Hash Algorithms (SHA and SHA-based HMAC and HKDF)", May 2011

Source of RFC: IETF - NON WORKING GROUP
Area Assignment: sec

Errata ID: 5240
Status: Reported
Type: Editorial
Publication Format(s) : TEXT
Reported By: Yeong-u Kim (K.)
Date Reported: 2018-01-21

Section 6 says: 

6.1.  SHA-224 and SHA-256 Initialization

   For SHA-224, the initial hash value, H(0), consists of the following
   32-bit words in hex:

      H(0)0 = c1059ed8
      H(0)1 = 367cd507
      H(0)2 = 3070dd17
      H(0)3 = f70e5939
      H(0)4 = ffc00b31
      H(0)5 = 68581511
      H(0)6 = 64f98fa7
      H(0)7 = befa4fa4

It should say:

6.1.  SHA-224 and SHA-256 Initialization

   For SHA-224, the initial hash value, H(0), consists of the following
   32-bit words in hex.  These words were obtained by taking the second
   32 bits of the fractional parts of the square roots of the ninth
   through sixteenth prime numbers.

      H(0)0 = c1059ed8
      H(0)1 = 367cd507
      H(0)2 = 3070dd17
      H(0)3 = f70e5939
      H(0)4 = ffc00b31
      H(0)5 = 68581511
      H(0)6 = 64f98fa7
      H(0)7 = befa4fa4

Notes:

The explanation for the way in which the initial hash value of SHA-224 was obtained is missing.

Errata ID: 6523
Status: Reported
Type: Editorial
Publication Format(s) : TEXT
Reported By: David Óscar Solé González
Date Reported: 2021-04-08

Section 6.4 says: 

   For i = 1 to N

      1. Prepare the message schedule W:
         For t = 0 to 15
            Wt = M(i)t
         For t = 16 to 79
            Wt = SSIG1(W(t-2)) + W(t-7) + SSIG0(W(t-15)) + W(t-16)

      2. Initialize the working variables:
         a = H(i-1)0
         b = H(i-1)1
         c = H(i-1)2
         d = H(i-1)3
         e = H(i-1)4
         f = H(i-1)5
         g = H(i-1)6
         h = H(i-1)7

      3. Perform the main hash computation:
         For t = 0 to 79
            T1 = h + BSIG1(e) + CH(e,f,g) + Kt + Wt
            T2 = BSIG0(a) + MAJ(a,b,c)
            h = g
            g = f
            f = e
            e = d + T1
            d = c
            c = b
            b = a
            a = T1 + T2

         4. Compute the intermediate hash value H(i)
            H(i)0 = a + H(i-1)0
            H(i)1 = b + H(i-1)1
            H(i)2 = c + H(i-1)2
            H(i)3 = d + H(i-1)3
            H(i)4 = e + H(i-1)4
            H(i)5 = f + H(i-1)5
            H(i)6 = g + H(i-1)6
            H(i)7 = h + H(i-1)7

It should say:

   For i = 1 to N

      1. Prepare the message schedule W:
         For t = 0 to 15
            Wt = M(i)t
         For t = 16 to 79
            Wt = SSIG1(W(t-2)) + W(t-7) + SSIG0(W(t-15)) + W(t-16)

      2. Initialize the working variables:
         a = H(i-1)0
         b = H(i-1)1
         c = H(i-1)2
         d = H(i-1)3
         e = H(i-1)4
         f = H(i-1)5
         g = H(i-1)6
         h = H(i-1)7

      3. Perform the main hash computation:
         For t = 0 to 79
            T1 = h + BSIG1(e) + CH(e,f,g) + Kt + Wt
            T2 = BSIG0(a) + MAJ(a,b,c)
            h = g
            g = f
            f = e
            e = d + T1
            d = c
            c = b
            b = a
            a = T1 + T2

      4. Compute the intermediate hash value H(i)
         H(i)0 = a + H(i-1)0
         H(i)1 = b + H(i-1)1
         H(i)2 = c + H(i-1)2
         H(i)3 = d + H(i-1)3
         H(i)4 = e + H(i-1)4
         H(i)5 = f + H(i-1)5
         H(i)6 = g + H(i-1)6
         H(i)7 = h + H(i-1)7

Notes:

Identation correction on point 4. as it may lead to confussion and reader may think that point (4) may be done inside a loop (begining on point (3).

Status: Held for Document Update (1)

RFC 6234, "US Secure Hash Algorithms (SHA and SHA-based HMAC and HKDF)", May 2011

Source of RFC: IETF - NON WORKING GROUP
Area Assignment: sec

Errata ID: 5238
Status: Held for Document Update
Type: Editorial
Publication Format(s) : TEXT
Reported By: Yeong-u Kim (K.)
Date Reported: 2018-01-19
Held for Document Update by: Warren Kumari (Ops AD)
Date Held: 2018-01-22

Section 3 says: 

   e. The rotate left (circular left shift) operation ROTL^n(x), where x
      is a w-bit word and n is an integer with 0 <= n < w, is defined by

         ROTL^n(X) = (x<<n) OR (x>>(w-n))

It should say:

   e. The rotate left (circular left shift) operation ROTL^n(x), where x
      is a w-bit word and n is an integer with 0 <= n < w, is defined by

         ROTL^n(x) = (x<<n) OR (x>>(w-n))

Notes:

A typo. The function argument has to be x, not X.

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